Re: Question on exact sequence

[Marco Grandis <grandis@dima.unige.it>]

Dear Michael,

The following lemma extends both results.

We have a sequence of consecutive morphisms indexed on the integers

	...  ---->   An  ---->  An+1  ---->  An+2  ---->  ...

(if your sequence is finite, you extend by zero's). Call  f(n,m)  the
composite
from  An  to  Am  (n < m).

Then, writing  H/K  a subquotient  H/(H intersection K),
there is an unbounded exact sequence of induced morphisms:

    ...  ---->   Ker f(n,n+2) / Im f(n-1,n)
    	---->   Ker f(n+1,n+2) / Im f(n-1,n+1)
    	---->   Ker f(n+1,n+3) / Im f(n,n+1) ---->  ...

where morphisms are alternatively induced by an 'elementary' morphism
(say An  -->  An+1) or by an identity.

At each step, one increases of one unit the first index in the numerator
and the second index in the denominator, or the opposite
(alternatively); after
two steps, all indices are increased of one unit, and we go along in
the same way.

-  Your lemma comes out of a sequence  A ----> B ----> C  (extended
with zeros).

-  Snake's lemma, with your letters, comes out of a sequence of three
morphisms
whose total composite is 0

      					 A ----> B ----> B' ----> C
taking into account that  A' = Ker(B' ----> C')  and  C = Cok(A ---->
B).

I like your lemma (and the Snake's). The form above does not look
really nice.
Perhaps someone else will find a nicer solution?

However, if one looks at the universal model of a sequence of
consecutive morphisms,
in my third paper on Distributive Homological Algebra, Cahiers 26,
1985, p.186,
the exact sequence above is obvious. (Much in the same way as for the
sequence of
the Snake Lemma, p. 188, diagrams (10) and (11).) This is how I found
it.

Best wishes

Marco




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